This answer on Math Overflow points out that

For instance, the fact that most of the mass of a unit ball in high dimensions lurks near the boundary of the ball can be interpreted as a manifestation of the law of large numbers, using the interpretation of a high-dimensional vector space as the state space for a large number of trials of a random variable.

I can hardly make any sense of it.

Anyway, here is my attempt: instead of addressing an $N$-dimensional unit ball, let's consider an $N$-dimensional unit cube first. Each point on that cube can be expressed with a coordinate consisting of $N$ numbers, i.e. $(X_1, X_2, ..., X_N)$, with $X_i \stackrel{iid}{\sim} \text{Uniform}[-1, 1]$. This independence is a result of the orthogonality of axes.

It is trivial to show that $E(X_i^2) = \frac{1}{3}$. Applying the Law of Large Numbers, the (euclidean) distance from a point $P$ to $\mathbb{O}$ is then

$$ \begin{equation} \begin{aligned} d(P, \mathbb{O}) &\stackrel{\text{def}}{=} \sqrt{X_1^2 + X_2^2 + ... + X_N^2} \\ E\left(d(P, \mathbb{O})^2\right) &= E\left(X_1^2 + X_2^2 + ... + X_N^2\right) \\ &\rightarrow \frac{N}{3} \end{aligned} \end{equation} $$

Recall that we are interested in the unit ball, so we have to somehow calculate the distribution of $d(P, \mathbb{O})^2 | d(P, \mathbb{O})^2 \le 1$, but I'm lost here.

My question is, how do you interpret the mass distribution of an $N$-dimensional unit ball with LLN?

  • $\begingroup$ +1 for raising the question. But you cannot interchange the expectation with a nonlinear operator. $E\left(d(P, \mathbb{O})\right) < \sqrt{\frac{N}{3}}$. $\endgroup$ – Hans May 27 '19 at 18:57
  • $\begingroup$ @Hans Thanks for pointing that out! I've edited the post to use $d^2$ instead. $\endgroup$ – nalzok May 28 '19 at 2:24
  • $\begingroup$ It seems to me LLN is not the right metaphor/whatever. The point is that if you have a large number of iid random variables uniformly distributed in ${-1,1}$, the probability that at least one of the random variables is within $\varepsilon$ of either -1 or 1 increases towards 1 as N goes to infinity. $\endgroup$ – pseudocydonia May 28 '19 at 4:07
  • $\begingroup$ @pseudocydonia I see your point, but how is this related to a high-dimensional unit ball? $\endgroup$ – nalzok May 28 '19 at 5:38
  • $\begingroup$ It applies most directly to the unit n-cube. $\endgroup$ – pseudocydonia May 28 '19 at 6:06

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